Analysis of formula for selecting $k$ elements from $n$ elements arranged in a row so that no two selected elements are consecutive 
How many ways to get $k$ elements in $n$ elements which are arranged in a line such that there does not exist two adjacent elements taken at the same time?

In the textbook I read, it is solved as follow:

As $k$ elements has been taken, there are $n - k$ remaining elements. There are $n - k + 1$ space among $n - k$ elements (which includes head and tail space). For each way to get $k$ spaces from these space will respond to one way to get $k$ elements satisfying the problem. Therefore, the result is (n - k + 1) C k

Actually I do not understand the way to solve as the guide I have mentioned above. I really need some one to explain more or give me another way to solve.
 A: Perhaps a concrete example will help.  Suppose we wish to select four numbers from the list $1, 2, 3, 4, 5, 6, 7, 8, 9, 10$ so that no two of the numbers are consecutive.
We will arrange six blue and four green balls so that no two of the green balls are consecutive.  First line up the six blue balls.  This creates seven spaces, five between successive blue balls and two at the ends of the row, as shown below.
$$\square b \square b \square b \square b \square b \square b \square$$
To ensure that no two of the green balls are consecutive, we must choose four of these seven spaces in which to place a green ball, which we can do in
$$\binom{7}{4} = \binom{10 - 4 + 1}{4}$$
ways.  For instance, if we pick the first, third, fourth, and seventh spaces, we will obtain the arrangement
$$gbbgbgbbbg$$
Now number the balls from left to right.  The numbers on the green balls are the desired set of four numbers, no two of which are consecutive.  In the arrangement above, the numbers on the green balls would be $1, 4, 6, 10$.
In the general case, we would have $n - k$ blue balls and $k$ green balls.  Lining up the $n - k$ blue balls creates $n - k + 1$ spaces, $n - k - 1$ spaces between successive blue balls and two at the ends of the row.  To ensure that no two of the green balls are consecutive, we must place a green ball in $k$ of these $n - k + 1$ spaces, which can be done in
$$\binom{n - k + 1}{k}$$
ways.  The positions of the green balls represent the positions of the objects which are selected so that no two consecutive objects are selected.
